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3 years ago

How many laws are named after sir issac newton

Physics

1 answer:

Blizzard [7]3 years ago

3 0

Answer:

Three Laws

Newton's Three Laws of Motion. Sir Isaac Newton: The Universal Law of Gravitation. Sir Isaac Newton and the Unification of Physics & Astronomy.

Explanation:

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Because of changes over time, the most accurate weather forecasts are: A) analog forecasts. B) long-term forecasts. C) seven-day

blondinia [14]

The answer is analog forecasts

8 0

3 years ago

Read 2 more answers

Using newtons law How much force is accelerated a 66kg skier 2m/s2

Evgen [1.6K]

Answer:

SUPONIENDO QUE NO HAY FRICCIÓN

Explanation:

F = 66 kg( $2 m/s^{2}$ ) = 132 N

8 0

3 years ago

A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and

iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

$a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g$

The speed at this point would then be

$v^2 = a_c r = 2gr = 2*9.8*17 = 333.2$

$v = \sqrt{333.2} = 18.25 m/s$

2. Similarly, if T = mg, then the centripetal acceleration must be

$a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0$

As the ball has no centripetal acceleration, its speed must also be 0 as well.

6 0

4 years ago

A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.

Alja [10]

Answer:

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

$Q = nC_p \Delta T$

for diatomic gas we have

$C_p = \frac{7}{2} R$

so we will have

$Q = 0.200(\frac{7}{2}R)(160 - 320)$

$Q = -930.7 J$

Now in adiabatic process heat is not transferred

so in this process

$Q = 0$

so we have

$T_1V_1^{1.4-1} = T_2V_2^{1.4-1}$

$(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}$

$T_2 = 121.26 K$

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

$Q_3 = nC_v\Delta T$

here we know that

$C_v = \frac{5}{2}R$

$Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)$

$Q_3 = 825.76 J$

So total heat given to the system is

$Q = -930.7 + 0 + 825.76$

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

3 0

3 years ago

Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto

timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

Vector B = ?

Resultant of vector A and B = R = -14j

R = A + B

-14j = 29j + B

B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

4 0

3 years ago